1. ## sum of series

Find the sum of the first n terms in the series:

(i) 1.3.5 + 3.5.7 + 5.7.9 + ... + (2n - 1)(2n + 1)(2n + 3)
(ii) 1/(1.3.5) + 1/(3.5.7) + 1/(5.7.9) + ... + 1/[(2n - 1)(2n + 1)(2n + 3)]

2. Hello, cedricc!

The first one is rather straight-forward.

Find the sum of the first $\displaystyle n$ terms in the series:

. . $\displaystyle (i)\;\;1\cdot3\cdot5 + 3\cdot5\cdot7 + 5\cdot7\cdot9 + \cdots + (2n - 1)(2n + 1)(2n + 3)$

We have: .$\displaystyle \sum^n_{k=1} (2k-1)(2k+1)(2k+3)$

. . . . . . . $\displaystyle =\; \sum^n_{k=1}\left(8k^3 + 12k^2 - 2k - 3\right)$

. . . . . . . $\displaystyle =\;8\sum^n_{k=1} k^3 \;+\; 12\sum^n_{k=1} k^2 \;-\; 2\sum^n_{k=1} k \;-\; 3\sum^n_{k=1} 1$

. . . . . . . $\displaystyle =\;8\cdot\frac{n^2(n+1)^2}{4} \;+\; 12\cdot\frac{n(n+1)(2n+1)}{6} \;-\; 2\cdot\frac{n(n+1)}{2} \;-\; 3n$

. . . . . . . $\displaystyle =\; 2n^4 + 8n^3 + 7n^2 - 2n$

. . . . . . . $\displaystyle =\;n(n+1)(2n^2+4n-1)$

3. Hi

For the second one

$\displaystyle \sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)(2k+3)} = \frac18 \sum_{k=1}^{n} \left(\frac{1}{2k-1}-\frac{2}{2k+1}+\frac{1}{2k+3}\right)$

Most of the terms cancel out

$\displaystyle \sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)(2k+3)} = \frac18 \left(\frac11 + \frac13-\frac23-\frac{1}{2n+1}+\frac{1}{2n+3}\right)$

After simplifications

$\displaystyle \sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)(2k+3)} = \frac{n(n+2)}{3(2n+1)(2n+3)}$